Difficulty: Intermediate | Prerequisites: Basic probability rules, combinatorics (factorials, combinations), expected value.
This set of distributions sits at the heart of an intro probability course. You have already met the Bernoulli trial (success/fail, fixed p). Every distribution here is built from repeated Bernoulli trials or from counting events, so if Bernoulli trials feel shaky, revisit those first. The geometric, negative binomial, Poisson and multinomial distributions each answer a different flavour of the same question: how many times does something happen, or how long until it does?
Four discrete distributions, each counting successes or trials in a different way. The geometric counts trials until the first success. The negative binomial counts trials until a set number of successes. The Poisson counts how many events land in a fixed window. The multinomial extends the binomial to more than two outcome categories. Know which one to reach for and you can set up most introductory discrete-probability problems in seconds.
In a first probability course, you typically start with the binomial: a fixed number of independent trials, each with the same probability of success. The distributions covered here are its relatives. The geometric and negative binomial flip the question (trials are random, successes are fixed). The Poisson handles events arriving at a constant average rate. The multinomial generalises the binomial so each trial can land in more than two buckets. Together they cover the standard toolkit for modelling count data and waiting times in discrete settings.
Geometric distribution
Models the number of independent Bernoulli trials needed to get the first success. Parameter: p (probability of success on each trial). Notation: X ~ Geometric(p).
Think of it as: "How many arrows do I shoot before I hit the bullseye?"
Negative binomial distribution
Models the number of trials needed to achieve a specified number of successes, r. Parameters: r (target successes), p (probability of success per trial). Notation: Y ~ NegBin(r, p).
In simple terms, this means: "How many trials until I collect r wins?" The geometric is the special case where r = 1.
Poisson distribution
Models the number of events occurring in a fixed interval (time, space, volume) when events arrive independently at a constant average rate. Parameter: lambda (the average number of events per interval). Notation: X ~ Poisson(lambda).
Think of it as: counting rare-ish events in a window, like misrecorded data entries per batch or customer arrivals per hour.
Multinomial distribution
Generalises the binomial to experiments where each trial can fall into k > 2 categories, each with its own probability. Parameters: n (number of trials), p1, p2, ..., pk (category probabilities summing to 1). Notation: (X1, ..., Xk) ~ Multinomial(n, p1, ..., pk).
In simple terms, this means: instead of just success/fail, each trial lands in one of several buckets. You are counting how many land in each bucket.
PMF (probability mass function)
The function that gives the probability of each possible value of a discrete random variable. Written P(X = x).
Think of it as: the rule that tells you exactly how likely each outcome is.
Expected value, E(X)
The long-run average of a random variable over many repetitions. For a geometric: E(X) = 1/p. For a negative binomial: E(Y) = r/p. For a Poisson: E(X) = lambda.
In simple terms, this means: if you repeated the experiment thousands of times, E(X) is the average you would converge on.
Setup and PMF
X = number of trials until the first success
Each trial is independent with constant success probability p
PMF: P(X = k) = (1 - p)^(k-1) * p, for k = 1, 2, 3, ...
The (1 - p)^(k-1) part counts the failures before the first success; the final p is the success itself
Expected value and variance
E(X) = 1/p
Var(X) = (1 - p) / p^2
Worked example (Problem 3.2.5)
An archer hits the bullseye with probability p = 0.09 on each independent shot. X ~ Geometric(0.09).
P(X = 4): the first bullseye lands on the 4th arrow.
P(X = 4) = (1 - 0.09)^3 * 0.09 = (0.91)^3 * 0.09 = 0.068
E(X) = 1/0.09 = 11.11 arrows on average before the first bullseye
Key intuition
The geometric distribution is memoryless: given that you have already failed k times, the probability of needing at least one more trial is exactly the same as it was at the start. No "momentum" builds up.
Setup and PMF
X = number of events in a fixed interval
Events occur independently at a constant average rate lambda
PMF: P(X = k) = (e^(-lambda) * lambda^k) / k!, for k = 0, 1, 2, ...
lambda is both the mean and the variance
Expected value and variance
E(X) = lambda
Var(X) = lambda
Worked example (Problem 3.4.8)
X ~ Poisson(9.2), where X is the number of misrecorded pieces of information.
P(6 <= X <= 10): sum the PMF from k = 6 to k = 10.
P(6 <= X <= 10) = sum of (e^(-9.2) * 9.2^k) / k! for k = 6, 7, 8, 9, 10 = 0.598
P(X <= 4): sum the PMF from k = 0 to k = 4.
P(X <= 4) = sum of (e^(-9.2) * 9.2^k) / k! for k = 0, 1, 2, 3, 4 = 0.0486
Key intuition
The Poisson is the go-to distribution when you are counting events that occur "randomly" in time or space with a known average rate. If someone says "on average 9.2 errors per batch," that is a Poisson setup. Large lambda values make the Poisson look roughly normal (symmetric, bell-shaped), while small lambda values produce a right-skewed shape.
Setup and PMF
Y = number of trials needed to achieve the r-th success
Each trial is independent with constant success probability p
PMF: P(Y = k) = C(k-1, r-1) * p^r * (1-p)^(k-r), for k = r, r+1, r+2, ...
The combination C(k-1, r-1) counts the ways to place the first (r-1) successes among the first (k-1) trials (the k-th trial is always the r-th success)
Expected value and variance
E(Y) = r/p
Var(Y) = r(1-p) / p^2
Connection to the geometric
The geometric distribution is the negative binomial with r = 1. If you understand the geometric, you already understand the building block: the negative binomial is the sum of r independent geometric random variables.
Worked examples
Problem 3.2.5 (part b): Y = number of arrows before the 3rd bullseye, with p = 0.09, so Y ~ NegBin(3, 0.09).
P(Y = 10) = C(9, 2) * (0.09)^3 * (0.91)^7 = 0.014
E(Y) = 3/0.09 = 33.33 arrows on average to get three bullseyes
Problem 3.8.7: A golf shop sells right-handed clubs (p = 0.42) and left-handed clubs (p = 0.58). X_i^R ~ NegBin(i, 0.42) and X_i^L ~ NegBin(j, 0.58).
P(3rd left-handed customer is the 4th customer) = C(8, 2) * (0.58)^4 * (0.42)^6 = 0.0299
P(4 <= X <= 7) for right-handed sales before 4 sets sold: sum the negative binomial PMF for the relevant values. Result: 0.329
Setup and PMF
n independent trials, each landing in one of k categories
Category i has probability p_i, with p_1 + p_2 + ... + p_k = 1
PMF: P(X_1 = x_1, ..., X_k = x_k) = n! / (x_1! * x_2! * ... * x_k!) * p_1^x_1 * p_2^x_2 * ... * p_k^x_k
Constraint: x_1 + x_2 + ... + x_k = n (all trials must be accounted for)
Expected value (per category)
E(X_i) = n * p_i
Var(X_i) = n * p_i * (1 - p_i)
Connection to the binomial
The binomial is the multinomial with k = 2 (success and failure). Each individual X_i in a multinomial is marginally Binomial(n, p_i), but the X_i values are not independent of each other because they must sum to n.
Worked example (Problem 3.5.6)
Five survey reports, each equally likely to be selected, so p_i = 0.2 for i = 1, 2, 3, 4, 5. Ten surveys distributed: n = 10.
P(X_1 = 2, X_2 = 2, X_3 = 2, X_4 = 2, X_5 = 2) = 10! / (2!)^5 * (1/5)^10 = 0.012
This is the probability that in 10 surveys, each of the five reports is selected exactly twice
Worked example (Problem 4.1.3)
X ~ Uniform(0,1). The unit interval is split into four sub-intervals: [0, 0.3), [0.3, 0.5), [0.5, 0.75), [0.75, 1]. Twenty numbers generated, so Y_1 through Y_4 count how many fall in each interval.
p_1 = 0.3, p_2 = 0.2, p_3 = 0.25, p_4 = 0.25
E(Y_1) = 20 * 0.3 = 6, Var(Y_1) = 20 * 0.3 * 0.7 = 4.2
E(Y_2) = 4, Var(Y_2) = 3.2
E(Y_3) = 5, Var(Y_3) = 3.75
E(Y_4) = 5, Var(Y_4) = 3.75
P(Y_1 = 5, Y_2 = 5, Y_3 = 5, Y_4 = 5) = 20! / (5!)^4 * 0.3^5 * 0.2^5 * 0.25^5 * 0.25^5 = 0.0089
Distribution | PMF | E(X) | Var(X) |
|---|---|---|---|
Geometric(p) | (1-p)^(k-1) * p | 1/p | (1-p)/p^2 |
NegBin(r, p) | C(k-1, r-1) * p^r * (1-p)^(k-r) | r/p | r(1-p)/p^2 |
Poisson(lambda) | e^(-lambda) * lambda^k / k! | lambda | lambda |
Multinomial(n, p_1...p_k) | n! / (x_1!...x_k!) * p_1^x_1...p_k^x_k | n*p_i (each) | np_i(1-p_i) (each) |
Geometric: Quality control on a production line. If the defect rate is 2%, the geometric distribution tells you how many items you expect to inspect before finding the first defective one.
Negative binomial: Clinical trials where you need a fixed number of treatment successes before stopping enrolment.
Poisson: Call centres model incoming calls per hour as Poisson to decide how many operators to staff. Insurance companies model the number of claims per month this way.
Multinomial: Market research surveys where respondents choose among several brands. The multinomial gives the probability of any particular split of responses across the options.
Students often confuse the geometric and the negative binomial. If the problem says "first success," it is geometric. If it says "r-th success" (with r > 1), it is negative binomial.
Students sometimes forget that the geometric PMF starts at k = 1 (you need at least one trial), not k = 0. Some textbooks define it starting at k = 0 (counting failures before the first success), which shifts the formulas. Check which convention your course uses.
The Poisson parameter lambda is both the mean and the variance. Students often treat these as separate quantities to be solved for independently. They are the same number.
In the multinomial, each individual X_i looks binomial on its own, but the X_i values are not independent. Students sometimes compute P(X_1 = a) * P(X_2 = b) as though they were. Use the joint PMF instead.
⚠️ Expect to be asked: "Which distribution models this scenario?" Practise identifying the distribution from a word problem before computing anything.
⚠️ The relationship between geometric and negative binomial is commonly tested. Know that Geometric = NegBin(1, p).
⚠️ For the Poisson, you will likely need to compute cumulative probabilities (summing the PMF over a range). Practise both P(X <= k) and P(a <= X <= b) by hand or with tables.
⚠️ Multinomial problems often test whether you can identify the correct n, k, and probabilities from a word problem. The factorial coefficient is the part students most often miscalculate.
True or false: The geometric distribution can take the value 0. (False, under the "trials until first success" convention; the minimum is 1.)
Fill in the blank: For a Poisson distribution, the variance equals ______. (lambda, the same as the mean.)
True or false: The negative binomial with r = 1 is the same as the geometric. (True.)
Fill in the blank: In a multinomial distribution with n trials, X_1 + X_2 + ... + X_k must equal ______. (n.)
True or false: In a multinomial, each X_i is independently distributed. (False; they are marginally binomial but not independent because they must sum to n.)
Q: A machine produces defective parts with probability 0.05. What is the probability that the first defective part is the 6th one inspected?
A: X ~ Geometric(0.05). P(X = 6) = (0.95)^5 * 0.05 = 0.0387.
Q: Using the same machine, what is the expected number of parts inspected before finding the first defective?
A: E(X) = 1/0.05 = 20 parts.
Q: A call centre receives an average of 12 calls per hour. What distribution would you use to model the number of calls in a given hour, and what is E(X)?
A: X ~ Poisson(12). E(X) = 12.
Q: You roll a fair die 18 times. What is the probability that each face appears exactly 3 times?
A: Multinomial with n = 18, k = 6, p_i = 1/6 for all i. P = 18! / (3!)^6 * (1/6)^18 = 0.0034 (approx.).
Q: A basketball player has a free-throw success rate of 0.80. What is the probability it takes exactly 5 attempts to make 3 free throws?
A: Y ~ NegBin(3, 0.80). P(Y = 5) = C(4, 2) * (0.80)^3 * (0.20)^2 = 6 * 0.512 * 0.04 = 0.1229.
These distributions connect directly to the binomial distribution, which you should already know. The exponential distribution (covered in the companion continuous distributions notes) is the continuous analogue of the geometric: it models waiting time until the first event in a Poisson process, just as the geometric models the number of discrete trials until the first success. The Poisson itself arises as a limit of the binomial when n is large and p is small (the "law of rare events").
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